Kinematics: Speed-Solving Techniques (10)

easy 2 min read

Question

A car starts from rest and accelerates uniformly at 4 m/s24\ \text{m/s}^2 for 55 seconds. Without writing the full equations, find the distance covered using a speed trick suitable for JEE/NEET MCQs.

Solution — Step by Step

Whenever the body starts from rest with uniform acceleration, the distance in time tt is simply half the area of the velocity-time triangle. We don’t need s=ut+12at2s = ut + \tfrac{1}{2}at^2 if we can read the geometry.

v=at=4×5=20 m/sv = at = 4 \times 5 = 20\ \text{m/s}. This is the height of the triangle.

s=12×base×height=12×5×20=50 ms = \tfrac{1}{2} \times \text{base} \times \text{height} = \tfrac{1}{2} \times 5 \times 20 = 50\ \text{m}

s=12(4)(25)=50 ms = \tfrac{1}{2}(4)(25) = 50\ \text{m}. Same answer in one line.

Final answer: s=50 ms = 50\ \text{m}.

Why This Works

Kinematics MCQs reward students who treat velocity-time graphs as the primary tool. Once you internalise that the area under a vv-tt graph is displacement and the slope is acceleration, problems collapse from three lines of algebra into one geometric step.

For JEE Main and NEET, almost every “from rest” question is a triangle. Every “constant velocity throw” question is a rectangle. Mixed motions become trapeziums. The trick is pattern recognition before formula recall.

Alternative Method

Use Galileo’s odd-number rule: for uniform acceleration from rest, the distances in the 1st, 2nd, 3rd… seconds are in the ratio 1:3:5:7:91:3:5:7:9. Sum of first 5 odd numbers is 25. With s1=12a(1)2=2 ms_1 = \tfrac{1}{2}a(1)^2 = 2\ \text{m}, total =25×2=50 m= 25 \times 2 = 50\ \text{m}.

For “ratio of distances” PYQs, jump straight to the odd-number rule. Saves 30 seconds per question.

Common Mistake

Students plug u=0u = 0 into s=ut+12at2s = ut + \tfrac{1}{2}at^2 and waste 20 seconds simplifying. When you see “starts from rest”, skip uu entirely and write s=12at2s = \tfrac{1}{2}at^2 in one stroke.

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